# Thread: True or false analysis.

1. ## True or false analysis.

Which of the following are true and which false? Give proofs or counterexamples.
i). If $x \rightarrow f(x)^2$ (this is a mapping, I couldn't quite find the right arrow) is continous at c then f is differentiable at c.
I said this was false and my counterexample was $f(x)=| \sqrt{x}|$.

This function is continous at 0 but $f(x)^2=|x|$ which is not differentiable at 0.

ii). If $\exists \ ga,b) \rightarrow \mathbb{R}" alt="\exists \ ga,b) \rightarrow \mathbb{R}" /> and $Za,b) \rightarrow \mathbb{R}" alt="Za,b) \rightarrow \mathbb{R}" /> such that $f(x)=g(x)Z(x)$ then f is differentiable at c if:

a). Z is bounded
b). g is differentiable at c with g'(c)=0 and
c). g(c)=0;

I think it's true but I don't know how to prove it.

iii). In ii) f is differentiable assuming only (a) and (c).
I put false and used the counterexample $f(x)=x \sin \left( \frac{1}{x} \right)$ when c=0.

$Z(x)= \sin \left( \frac{1}{x} \right)$ and $g(x)=x$.

In ii). f is differentiable assuming only (a) and (b).
Once again I put false and used the counterexample $f(x)=5 \sin \left( \frac{1}{x} \right)$ where c=0.

Where $g(x)=5$ and $Z(x)= \sin \left( \frac{1}{x} \right)$

In (ii). f is differentiable assuming only (b) and (c).
This is false because $Z(x)$ could be discontinous.

My counterexample was $f(x)=x^2. \begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\
0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}$
at c=0.

Here $g(x)=x^2$ and $Z(x)=\begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\
0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}$
.

are these correct?

2. Originally Posted by Showcase_22
I said this was false and my counterexample was $f(x)=| \sqrt{x}|$.

This function is continous at 0 but $f(x)^2=|x|$ which is not differentiable at 0.
You mean $f(x)=\sqrt{|x|}$ I suppose ?
If so it is OK

3. Originally Posted by Showcase_22

I think it's true but I don't know how to prove it.
Yes it is true

$\frac{f(x) - f(c)}{x-c} = \frac{g(x)Z(x) - g(c)Z(c)}{x-c} = \frac{g(x)Z(x)}{x-c} = Z(x) \:\frac{g(x)}{x-c} = Z(x) \:\frac{g(x) - g(c)}{x-c}$

I think that you can finish.
This will give you ideas for the other cases.

4. Originally Posted by Showcase_22
I put false and used the counterexample $f(x)=x \sin \left( \frac{1}{x} \right)$ when c=0.

$Z(x)= \sin \left( \frac{1}{x} \right)$ and $g(x)=x$.
I do not agree with your counter-example since g is not defined in 0 (even if its limit is 0).

Take $Z(x) = 1 \mbox{ and } g(x) = \sqrt{|x|}$ over (-1,1) for instance

Z is bounded and g(0)=0 therefore (a) and (c) are verified

$f(x) = \sqrt{|x|}$ is not differentiable in 0